1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Use symmetry in double integral.

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the double integral of (2+xy^2) over dA (dxdy) using symmetry where R = [0,1] x [-1,1]



    2. Relevant equations



    3. The attempt at a solution
    I don't know how to use symmetry to evaluate this.. However if I integrate this integral normally
    i first get [2x+2yx^3/3] between 0 and 1. Then i'd take the integral of that between -1 and 1? Where does symmetry come into play? Thanks :D
     
  2. jcsd
  3. Oct 24, 2013 #2

    Mark44

    Staff: Mentor

    If you let f(x, y) = 2 + xy2, then f(x, -y) = f(x, y). This means that the function is symmetric across the x-z plane. However the graph appears over the upper half of region R is the mirror image of how it appears over the lower half.

    Your integral represents the volume of the solid whose base is your region R. The symmetry allows you to find the volume over the upper half ([0, 1] X [0, 1]) and double it.
     
  4. Oct 24, 2013 #3
    The only obvious symmetries to me are that the x reflects across the x-axis and the ##y^2## across the y axis. Using either you get ##\int \int(2 + xy^2) dA = 2\int_0^1 \int_0^1 (2 + xy^2)dxdy##; or you can do the same integral dydx, but you could anyway.

    Although I am a great fan of using symmetry to dodge work, I don't see how this helps you much. Perhaps there is some symmetry I don't see.
     
  5. Oct 25, 2013 #4

    Mark44

    Staff: Mentor

    ???

    The solid whose volume is being found has an upper surface of z = f(x, y) = 2 + xy2. Pretty clearly, f(x, -y) = f(x, y), which means there is symmetry across the x-z plane, as I said in my previous post.
     
  6. Oct 25, 2013 #5

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Mark, I think the point is that the new integral to evaluate is literally just as difficult to evaluate as the original one, unless you find calculating (-1)3 to be unusually hard. I don't see any way to reduce the integral more than that though, so I guess it's what they intend you to do.
     
  7. Oct 25, 2013 #6

    Mark44

    Staff: Mentor

    I agree. I think it's just an exercise to get the student to use symmetry, not that it saves a lot of work. That's my sense at any rate.
     
  8. Oct 25, 2013 #7
    I agree with both of you, and think this is not a useful example. There are many good examples where the symmetry would really help, so any students solving it would learn to keep an eye out for symmetries.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted